Question

If the line \(\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}\) lies in the plane \(x+3y+\alpha z+\beta =0\), then \((\alpha ,\beta )\) is

• A. \((6,7)\)
• B. \((-6,7)\)
• C. \((6,-7)\)
• D. \((-6,-7)\)

Hint:

The line direction must be perpendicular to the plane normal and a point of the line must satisfy the plane equation.

Answer 1

The Correct Option is A

Step 1: Two Conditions
A line lies in a plane if its direction is parallel to the plane and one of its points is on the plane.

Step 2: Use the Point on the Line
Substituting $(2,1,-2)$: $2+3-2\alpha+\beta=0$, so $\beta=2\alpha-5$.

Step 3: Use the Direction
$3-15+2\alpha=0$ gives $\alpha=6$, so $\beta=12-5=7$, option (A).

Final Answer:
The values are alpha = 6 and beta = 7. \[ \boxed{\text{(A) }(6,7)} \]